Samacheer Kalvi Class 11 Maths Solutions Chapter 2 Basic Algebra Exercise 2.11

Question 1. Simplify
(a) \( (125)^{2/3} \)
(b) \( 16^{-3/4} \)
(c) \( (-1000)^{-2/3} \)
(d) \( (3^{-6})^{1/3} \)
(e) \( \frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}} \)
Answer:
(a) For \( (125)^{2/3} \):
First, we know that \( 125 \) can be written as \( 5^3 \).
So, \( (125)^{2/3} = (5^3)^{2/3} \)
Using the power rule \( (a^m)^n = a^{m \times n} \), we multiply the exponents.
\( (5^3)^{2/3} = 5^{3 \times \frac{2}{3}} \)
\( = 5^2 \)
\( = 25 \)
Thus, \( (125)^{2/3} \) simplifies to 25. This uses the property of fractional exponents relating to roots and powers.

(b) For \( 16^{-3/4} \):
First, \( 16 \) can be written as \( 2^4 \).
So, \( 16^{-3/4} = (2^4)^{-3/4} \)
Using the power rule, \( (a^m)^n = a^{m \times n} \), we multiply the exponents.
\( (2^4)^{-3/4} = 2^{4 \times (-\frac{3}{4})} \)
\( = 2^{-3} \)
A negative exponent means we take the reciprocal: \( a^{-n} = \frac{1}{a^n} \).
\( 2^{-3} = \frac{1}{2^3} \)
\( = \frac{1}{8} \)
So, \( 16^{-3/4} \) simplifies to \( \frac{1}{8} \). Understanding negative and fractional exponents is key here.

(c) For \( (-1000)^{-2/3} \):
First, \( -1000 \) can be written as \( (-10)^3 \).
So, \( (-1000)^{-2/3} = ((-10)^3)^{-2/3} \)
Using the power rule, \( (a^m)^n = a^{m \times n} \), we multiply the exponents.
\( ((-10)^3)^{-2/3} = (-10)^{3 \times (-\frac{2}{3})} \)
\( = (-10)^{-2} \)
A negative exponent means we take the reciprocal: \( a^{-n} = \frac{1}{a^n} \).
\( (-10)^{-2} = \frac{1}{(-10)^2} \)
\( = \frac{1}{100} \)
Thus, \( (-1000)^{-2/3} \) simplifies to \( \frac{1}{100} \). Remember that squaring a negative number results in a positive number.

(d) For \( (3^{-6})^{1/3} \):
Using the power rule, \( (a^m)^n = a^{m \times n} \), we multiply the exponents.
\( (3^{-6})^{1/3} = 3^{-6 \times \frac{1}{3}} \)
\( = 3^{-2} \)
A negative exponent means we take the reciprocal: \( a^{-n} = \frac{1}{a^n} \).
\( 3^{-2} = \frac{1}{3^2} \)
\( = \frac{1}{9} \)
So, \( (3^{-6})^{1/3} \) simplifies to \( \frac{1}{9} \). This problem directly tests the power of a power rule.

(e) For \( \frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}} \):
We can rewrite \( 27 \) as \( 3^3 \).
So, \( \frac{(3^3)^{-\frac{2}{3}}}{(3^3)^{-\frac{1}{3}}} \)
Applying the power rule \( (a^m)^n = a^{m \times n} \) to the numerator and denominator:
Numerator: \( (3^3)^{-\frac{2}{3}} = 3^{3 \times (-\frac{2}{3})} = 3^{-2} \)
Denominator: \( (3^3)^{-\frac{1}{3}} = 3^{3 \times (-\frac{1}{3})} = 3^{-1} \)
So the expression becomes \( \frac{3^{-2}}{3^{-1}} \)
Using the quotient rule for exponents \( \frac{a^m}{a^n} = a^{m-n} \):
\( = 3^{-2 - (-1)} \)
\( = 3^{-2 + 1} \)
\( = 3^{-1} \)
A negative exponent means we take the reciprocal: \( a^{-n} = \frac{1}{a^n} \).
\( = \frac{1}{3} \)
Thus, \( \frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}} \) simplifies to \( \frac{1}{3} \). This requires knowing how to handle fractional and negative exponents, as well as exponent rules for division.
In simple words: To simplify expressions with powers and fractions, first change the base number into its prime factors. Then, use the rules for multiplying exponents, dividing exponents, and handling negative exponents. Remember that a negative exponent means you flip the number and make the exponent positive.

🎯 Exam Tip: Always break down the base numbers into their prime factors first to make calculations with fractional exponents easier. Remember the rules for exponents, especially for negative and fractional powers.

Question 2. Evaluate \( \left[\left(256\right)^{-\frac{1}{2}}\right]^{-\frac{1}{4}} \cdot \left[256^{\frac{1}{3}}\right]^{\frac{3}{8}} \)
Answer:
We need to evaluate the expression: \( \left[\left(256\right)^{-\frac{1}{2}}\right]^{-\frac{1}{4}} \cdot \left[256^{\frac{1}{3}}\right]^{\frac{3}{8}} \)
Let's simplify each part separately.
For the first part, \( \left[\left(256\right)^{-\frac{1}{2}}\right]^{-\frac{1}{4}} \):
Using the power rule \( (a^m)^n = a^{m \times n} \), we multiply the exponents.
\( (256)^{-\frac{1}{2} \times (-\frac{1}{4})} \)
\( = (256)^{\frac{1}{8}} \)
We know that \( 256 \) can be written as \( 2^8 \).
\( = (2^8)^{\frac{1}{8}} \)
Applying the power rule again:
\( = 2^{8 \times \frac{1}{8}} \)
\( = 2^1 \)
\( = 2 \)
For the second part, \( \left[256^{\frac{1}{3}}\right]^{\frac{3}{8}} \):
Using the power rule \( (a^m)^n = a^{m \times n} \), we multiply the exponents.
\( = 256^{\frac{1}{3} \times \frac{3}{8}} \)
\( = 256^{\frac{1}{8}} \)
Again, \( 256 \) can be written as \( 2^8 \).
\( = (2^8)^{\frac{1}{8}} \)
Applying the power rule:
\( = 2^{8 \times \frac{1}{8}} \)
\( = 2^1 \)
\( = 2 \)
Now, we multiply the simplified results of both parts:
First part \( \times \) Second part \( = 2 \times 2 = 4 \).
Therefore, the value of the entire expression is 4. It's useful to simplify exponents step-by-step to avoid errors.
In simple words: First, work out each part of the problem separately. When you see powers of powers, multiply the small numbers on top. Then, change the main number into its smallest possible base number (like 256 becomes \( 2^8 \)). After simplifying each part to a single number, multiply them together to get the final answer.

🎯 Exam Tip: Always simplify the base number to its prime factorization (e.g., \( 256 = 2^8 \)) early in the problem. This makes it much easier to multiply and divide fractional exponents without complex calculations.

Question 3. If \( \left(x^{\frac{1}{2}}+x^{-\frac{1}{2}}\right)^{2}=\frac{9}{2} \) then find the value of \( \left(x^{\frac{1}{2}}-x^{-\frac{1}{2}}\right) \) for \( x > 1 \).
Answer:
We are given the equation: \( \left(x^{\frac{1}{2}}+x^{-\frac{1}{2}}\right)^{2}=\frac{9}{2} \)
Let's expand the left side using the formula \( (a+b)^2 = a^2 + b^2 + 2ab \), where \( a = x^{\frac{1}{2}} \) and \( b = x^{-\frac{1}{2}} \).
\( \left(x^{\frac{1}{2}}\right)^2 + \left(x^{-\frac{1}{2}}\right)^2 + 2 \left(x^{\frac{1}{2}}\right)\left(x^{-\frac{1}{2}}\right) = \frac{9}{2} \)
Simplify the terms:
\( x^{\frac{1}{2} \times 2} + x^{-\frac{1}{2} \times 2} + 2 x^{\frac{1}{2} - \frac{1}{2}} = \frac{9}{2} \)
\( x^1 + x^{-1} + 2 x^0 = \frac{9}{2} \)
Remember that \( x^0 = 1 \) and \( x^{-1} = \frac{1}{x} \).
\( x + \frac{1}{x} + 2(1) = \frac{9}{2} \)
\( x + \frac{1}{x} + 2 = \frac{9}{2} \)
Now, subtract 2 from both sides to find \( x + \frac{1}{x} \).
\( x + \frac{1}{x} = \frac{9}{2} - 2 \)
\( x + \frac{1}{x} = \frac{9-4}{2} \)
\( x + \frac{1}{x} = \frac{5}{2} \) (Equation 1)

Next, we need to find the value of \( \left(x^{\frac{1}{2}}-x^{-\frac{1}{2}}\right) \).
Let's consider \( \left(x^{\frac{1}{2}}-x^{-\frac{1}{2}}\right)^2 \).
Using the formula \( (a-b)^2 = a^2 + b^2 - 2ab \), where \( a = x^{\frac{1}{2}} \) and \( b = x^{-\frac{1}{2}} \).
\( \left(x^{\frac{1}{2}}-x^{-\frac{1}{2}}\right)^2 = \left(x^{\frac{1}{2}}\right)^2 + \left(x^{-\frac{1}{2}}\right)^2 - 2 \left(x^{\frac{1}{2}}\right)\left(x^{-\frac{1}{2}}\right) \)
\( = x^1 + x^{-1} - 2 x^0 \)
\( = x + \frac{1}{x} - 2(1) \)
\( = x + \frac{1}{x} - 2 \)
Now, substitute the value of \( x + \frac{1}{x} \) from Equation 1 into this expression.
\( \left(x^{\frac{1}{2}}-x^{-\frac{1}{2}}\right)^2 = \frac{5}{2} - 2 \)
\( = \frac{5-4}{2} \)
\( = \frac{1}{2} \)
To find \( \left(x^{\frac{1}{2}}-x^{-\frac{1}{2}}\right) \), we take the square root of both sides.
\( x^{\frac{1}{2}}-x^{-\frac{1}{2}} = \pm \sqrt{\frac{1}{2}} \)
\( = \pm \frac{1}{\sqrt{2}} \)
Since the problem states \( x > 1 \), this means \( x^{\frac{1}{2}} \) will be greater than 1, and \( x^{-\frac{1}{2}} = \frac{1}{x^{\frac{1}{2}}} \) will be less than 1. Therefore, \( x^{\frac{1}{2}}-x^{-\frac{1}{2}} \) must be positive.
So, \( x^{\frac{1}{2}}-x^{-\frac{1}{2}} = \frac{1}{\sqrt{2}} \). This is a good example of how given conditions like \( x > 1 \) are important for the final answer.
In simple words: First, use the given equation to find the value of \( x + \frac{1}{x} \). Then, use a known algebraic rule to find \( \left(x^{\frac{1}{2}}-x^{-\frac{1}{2}}\right)^2 \) in terms of \( x + \frac{1}{x} \). Finally, take the square root and choose the positive answer because the problem says \( x \) is greater than 1.

🎯 Exam Tip: When dealing with expressions like \( (a+b)^2 \) and \( (a-b)^2 \), remember their expanded forms. Also, always check conditions like \( x > 1 \) to determine if the positive or negative square root is the correct answer.

Question 4. Simplify and hence find the value of n: \( \frac{3^{2 n} \cdot 9^{2} \cdot 3^{-n}}{3^{3 n}} = 27 \)
Answer:
We are given the equation: \( \frac{3^{2 n} \cdot 9^{2} \cdot 3^{-n}}{3^{3 n}} = 27 \)
First, let's rewrite all terms with a base of 3. We know that \( 9 = 3^2 \) and \( 27 = 3^3 \).
So, the equation becomes:
\( \frac{3^{2 n} \cdot (3^2)^{2} \cdot 3^{-n}}{3^{3 n}} = 3^3 \)
Apply the power rule \( (a^m)^n = a^{m \times n} \) to \( (3^2)^2 \).
\( \frac{3^{2 n} \cdot 3^{2 \times 2} \cdot 3^{-n}}{3^{3 n}} = 3^3 \)
\( \frac{3^{2 n} \cdot 3^{4} \cdot 3^{-n}}{3^{3 n}} = 3^3 \)
Now, use the product rule \( a^m \cdot a^n = a^{m+n} \) for the terms in the numerator.
\( \frac{3^{2 n + 4 - n}}{3^{3 n}} = 3^3 \)
\( \frac{3^{n + 4}}{3^{3 n}} = 3^3 \)
Next, use the quotient rule \( \frac{a^m}{a^n} = a^{m-n} \) for the left side.
\( 3^{(n+4) - 3n} = 3^3 \)
\( 3^{4 - 2n} = 3^3 \)
Since the bases are the same, we can equate the exponents.
\( 4 - 2n = 3 \)
Now, solve for n.
Subtract 4 from both sides:
\( -2n = 3 - 4 \)
\( -2n = -1 \)
Divide by -2:
\( n = \frac{-1}{-2} \)
\( n = \frac{1}{2} \)
The value of n is \( \frac{1}{2} \). This problem shows how important it is to convert all bases to the same number before applying exponent rules.
In simple words: First, change all the numbers in the equation to have the same base number, which is 3 in this case. Then, use the rules for adding and subtracting the small power numbers (exponents) when multiplying or dividing numbers with the same base. Once the bases on both sides are the same, you can just set the exponents equal to each other and solve for 'n'.

🎯 Exam Tip: When solving equations with exponents, always try to express all terms with the same base. This simplifies the problem significantly, allowing you to equate the exponents directly.

Question 5. Find the radius of the spherical tank whose volume is \( \frac{32 \pi}{3} \) units.
Answer:
Let 'r' be the radius of the spherical tank.
The formula for the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \).
We are given that the volume of the spherical tank is \( \frac{32 \pi}{3} \) units.
So, we can set up the equation:
\( \frac{4}{3} \pi r^3 = \frac{32 \pi}{3} \)
To solve for 'r', we can cancel out common terms from both sides.
Multiply both sides by 3 to clear the denominators:
\( 4 \pi r^3 = 32 \pi \)
Now, divide both sides by \( 4 \pi \):
\( r^3 = \frac{32 \pi}{4 \pi} \)
\( r^3 = 8 \)
To find 'r', we take the cube root of both sides.
\( r = \sqrt[3]{8} \)
We know that \( 8 = 2^3 \).
\( r = \sqrt[3]{2^3} \)
\( r = 2 \)
Thus, the radius of the spherical tank is 2 units. This uses a direct application of the volume formula for a sphere.
In simple words: We use the known formula for the volume of a sphere. We set this formula equal to the volume given in the problem. Then, we solve for the radius 'r' by canceling out similar parts and taking the cube root of the number that is left.

🎯 Exam Tip: Always remember the basic formulas for geometric shapes, such as the volume of a sphere (\( \frac{4}{3} \pi r^3 \)). Make sure to show each step of solving the equation clearly.

Question 6. Simplify by rationalizing the denominator \( \frac{7+\sqrt{6}}{3-\sqrt{2}} \)
Answer:
We need to simplify the expression \( \frac{7+\sqrt{6}}{3-\sqrt{2}} \) by rationalizing the denominator.
To rationalize a denominator of the form \( (a-b) \), we multiply both the numerator and the denominator by its conjugate, which is \( (a+b) \).
Here, the denominator is \( 3-\sqrt{2} \), so its conjugate is \( 3+\sqrt{2} \).
\( \frac{7+\sqrt{6}}{3-\sqrt{2}} = \frac{7+\sqrt{6}}{3-\sqrt{2}} \times \frac{3+\sqrt{2}}{3+\sqrt{2}} \)
Now, multiply the numerators and the denominators.
Numerator: \( (7+\sqrt{6})(3+\sqrt{2}) \)
\( = 7(3) + 7(\sqrt{2}) + \sqrt{6}(3) + \sqrt{6}(\sqrt{2}) \)
\( = 21 + 7\sqrt{2} + 3\sqrt{6} + \sqrt{12} \)
We can simplify \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \).
So, the numerator becomes \( 21 + 7\sqrt{2} + 3\sqrt{6} + 2\sqrt{3} \)

Denominator: \( (3-\sqrt{2})(3+\sqrt{2}) \)
This is in the form \( (a-b)(a+b) = a^2 - b^2 \).
\( = 3^2 - (\sqrt{2})^2 \)
\( = 9 - 2 \)
\( = 7 \)

Now, combine the simplified numerator and denominator.
\( \frac{21 + 7\sqrt{2} + 3\sqrt{6} + 2\sqrt{3}}{7} \)
This expression has a rational denominator. We can also write it as:
\( = \frac{21}{7} + \frac{7\sqrt{2}}{7} + \frac{3\sqrt{6}}{7} + \frac{2\sqrt{3}}{7} \)
\( = 3 + \sqrt{2} + \frac{3\sqrt{6}}{7} + \frac{2\sqrt{3}}{7} \)
The expression is now simplified with a rational denominator. Rationalizing helps remove irrational numbers from the bottom part of a fraction, making it easier to work with.
In simple words: To remove the square root from the bottom of the fraction, multiply the top and bottom by the "conjugate" of the denominator. This means if the bottom is \( (A-B) \), you multiply by \( (A+B) \). Then, expand the top and bottom parts. The bottom will become a simple number, and the top will be a sum of terms with square roots.

🎯 Exam Tip: Remember to multiply by the conjugate of the denominator for rationalization. The conjugate of \( a-\sqrt{b} \) is \( a+\sqrt{b} \), and when multiplied, it results in \( a^2 - b \).

Question 7. Simplify: \( \frac{1}{3-\sqrt{8}} - \frac{1}{\sqrt{8}-\sqrt{7}} + \frac{1}{\sqrt{7}-\sqrt{6}} - \frac{1}{\sqrt{6}-\sqrt{5}} + \frac{1}{\sqrt{5}-2} \)
Answer:
Let P be the given expression: \( P = \frac{1}{3-\sqrt{8}} - \frac{1}{\sqrt{8}-\sqrt{7}} + \frac{1}{\sqrt{7}-\sqrt{6}} - \frac{1}{\sqrt{6}-\sqrt{5}} + \frac{1}{\sqrt{5}-2} \)
We will rationalize each term separately. Remember that \( \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \).

Term 1: \( \frac{1}{3-\sqrt{8}} \)
Multiply by \( \frac{3+\sqrt{8}}{3+\sqrt{8}} \):
\( = \frac{1}{3-\sqrt{8}} \times \frac{3+\sqrt{8}}{3+\sqrt{8}} = \frac{3+\sqrt{8}}{3^2 - (\sqrt{8})^2} = \frac{3+\sqrt{8}}{9-8} = \frac{3+\sqrt{8}}{1} = 3+\sqrt{8} \) (Equation 1)

Term 2: \( \frac{1}{\sqrt{8}-\sqrt{7}} \)
Multiply by \( \frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}} \):
\( = \frac{1}{\sqrt{8}-\sqrt{7}} \times \frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}+\sqrt{7}} = \frac{\sqrt{8}+\sqrt{7}}{(\sqrt{8})^2 - (\sqrt{7})^2} = \frac{\sqrt{8}+\sqrt{7}}{8-7} = \frac{\sqrt{8}+\sqrt{7}}{1} = \sqrt{8}+\sqrt{7} \) (Equation 2)

Term 3: \( \frac{1}{\sqrt{7}-\sqrt{6}} \)
Multiply by \( \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}} \):
\( = \frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}} = \frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} = \frac{\sqrt{7}+\sqrt{6}}{7-6} = \frac{\sqrt{7}+\sqrt{6}}{1} = \sqrt{7}+\sqrt{6} \) (Equation 3)

Term 4: \( \frac{1}{\sqrt{6}-\sqrt{5}} \)
Multiply by \( \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}} \):
\( = \frac{1}{\sqrt{6}-\sqrt{5}} \times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}} = \frac{\sqrt{6}+\sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2} = \frac{\sqrt{6}+\sqrt{5}}{6-5} = \frac{\sqrt{6}+\sqrt{5}}{1} = \sqrt{6}+\sqrt{5} \) (Equation 4)

Term 5: \( \frac{1}{\sqrt{5}-2} \)
Multiply by \( \frac{\sqrt{5}+2}{\sqrt{5}+2} \):
\( = \frac{1}{\sqrt{5}-2} \times \frac{\sqrt{5}+2}{\sqrt{5}+2} = \frac{\sqrt{5}+2}{(\sqrt{5})^2 - 2^2} = \frac{\sqrt{5}+2}{5-4} = \frac{\sqrt{5}+2}{1} = \sqrt{5}+2 \) (Equation 5)

Now, substitute these simplified terms back into the original expression for P.
\( P = (3+\sqrt{8}) - (\sqrt{8}+\sqrt{7}) + (\sqrt{7}+\sqrt{6}) - (\sqrt{6}+\sqrt{5}) + (\sqrt{5}+2) \)
Remove the parentheses and distribute the negative signs:
\( P = 3+\sqrt{8} - \sqrt{8}-\sqrt{7} + \sqrt{7}+\sqrt{6} - \sqrt{6}-\sqrt{5} + \sqrt{5}+2 \)
Now, cancel out the matching positive and negative square root terms:
\( P = 3 + (\sqrt{8}-\sqrt{8}) + (-\sqrt{7}+\sqrt{7}) + (\sqrt{6}-\sqrt{6}) + (-\sqrt{5}+\sqrt{5}) + 2 \)
\( P = 3 + 0 + 0 + 0 + 0 + 2 \)
\( P = 5 \)
The simplified value of the expression is 5. This method of rationalizing each term individually and then summing them is very effective for such series.
In simple words: For each part of the sum, get rid of the square root on the bottom by multiplying the top and bottom by its "conjugate" (changing the minus to a plus, or vice-versa). After doing this for every part, many of the square root terms will cancel each other out, leaving only simple numbers. Add these simple numbers together to get the final answer.

🎯 Exam Tip: When simplifying a series of fractions with square roots in the denominator, always rationalize each term individually. This often creates a "telescoping sum" where intermediate terms cancel out, leading to a much simpler final result.

Question 8. If \( x = \sqrt{2} + \sqrt{3} \) find \( \frac{x^{2}+1}{x^{2}-2} \)
Answer:
We are given \( x = \sqrt{2} + \sqrt{3} \).
First, let's find \( x^2 \).
\( x^2 = (\sqrt{2} + \sqrt{3})^2 \)
Using the formula \( (a+b)^2 = a^2 + b^2 + 2ab \):
\( x^2 = (\sqrt{2})^2 + (\sqrt{3})^2 + 2(\sqrt{2})(\sqrt{3}) \)
\( x^2 = 2 + 3 + 2\sqrt{6} \)
\( x^2 = 5 + 2\sqrt{6} \)

Now, we need to find the expression \( \frac{x^{2}+1}{x^{2}-2} \).
Substitute the value of \( x^2 \) into the numerator \( x^2+1 \):
\( x^2+1 = (5 + 2\sqrt{6}) + 1 \)
\( = 6 + 2\sqrt{6} \)

Substitute the value of \( x^2 \) into the denominator \( x^2-2 \):
\( x^2-2 = (5 + 2\sqrt{6}) - 2 \)
\( = 3 + 2\sqrt{6} \)

So the expression becomes:
\( \frac{x^{2}+1}{x^{2}-2} = \frac{6 + 2\sqrt{6}}{3 + 2\sqrt{6}} \)
To simplify this fraction, we need to rationalize the denominator. Multiply the numerator and denominator by the conjugate of the denominator, which is \( 3 - 2\sqrt{6} \).
\( = \frac{6 + 2\sqrt{6}}{3 + 2\sqrt{6}} \times \frac{3 - 2\sqrt{6}}{3 - 2\sqrt{6}} \)

Numerator: \( (6 + 2\sqrt{6})(3 - 2\sqrt{6}) \)
\( = 6(3) + 6(-2\sqrt{6}) + 2\sqrt{6}(3) + 2\sqrt{6}(-2\sqrt{6}) \)
\( = 18 - 12\sqrt{6} + 6\sqrt{6} - 4(\sqrt{6})^2 \)
\( = 18 - 6\sqrt{6} - 4(6) \)
\( = 18 - 6\sqrt{6} - 24 \)
\( = -6 - 6\sqrt{6} \)

Denominator: \( (3 + 2\sqrt{6})(3 - 2\sqrt{6}) \)
This is in the form \( (a+b)(a-b) = a^2 - b^2 \).
\( = 3^2 - (2\sqrt{6})^2 \)
\( = 9 - (2^2 \times (\sqrt{6})^2) \)
\( = 9 - (4 \times 6) \)
\( = 9 - 24 \)
\( = -15 \)

Now, put the simplified numerator and denominator back into the fraction.
\( \frac{-6 - 6\sqrt{6}}{-15} \)
We can factor out -6 from the numerator:
\( = \frac{-6(1 + \sqrt{6})}{-15} \)
Cancel the common factor of -3:
\( = \frac{2(1 + \sqrt{6})}{5} \)
\( = \frac{2 + 2\sqrt{6}}{5} \)
So, the simplified value of the expression is \( \frac{2 + 2\sqrt{6}}{5} \). This problem combines squaring binomials with surds and then rationalizing the denominator of a fraction.
In simple words: First, square the given value of \( x \) to find \( x^2 \). Then, put this \( x^2 \) value into the fraction \( \frac{x^{2}+1}{x^{2}-2} \) and simplify the top and bottom. To remove the square root from the bottom of the fraction, multiply the top and bottom by the "conjugate" of the denominator. Finally, simplify the resulting fraction as much as possible.

🎯 Exam Tip: Remember to apply the algebraic identity \( (a+b)^2 = a^2+2ab+b^2 \) when squaring terms with square roots. For rationalization, use the conjugate \( (a-b)(a+b) = a^2-b^2 \) to eliminate square roots from the denominator.